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$$\left|\frac{x^2} {x-1}\right|\leq 1$$

In the case where $x>1$, there are no real roots.

So in the case where $x<1$:

I opened the L.H.S. with a $-$ sign (as the modulus function has to remain positive). After this I multiplied both sides of the inequation by $-1$ giving me: $\frac{x^2}{x-1} \geq -1$ [The inequality sign flips due to multiplication with a negative number].

What have I done wrong, as doing this led me to the wrong answer...

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    $\begingroup$ You did nothing wrong so far. $\endgroup$ – mathlove Dec 29 '16 at 10:08
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As you said $x>1$ is not possible so suppose $x<1$ then

$$ \left|\frac{x^2}{1-x}\right|=\frac{x^2}{1-x}\leq 1 $$

So if you rearrange this you get $x^2+x-1\leq 0$. Using quadratic formula you get $\frac{-1-1\sqrt{5}}{2}<x<\frac{-1+\sqrt{5}}{2}$

With what you are doing steps are following

$$ \frac{x^2}{x-1}\geq -1 $$

Multiply both sides with $x-1<0$ then inequality change its sign again and

$$ x^2\leq -x+1 $$ which yields the same.

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  • $\begingroup$ Thanks a lot! I realized the mistake. $\endgroup$ – Harsha G. Jan 1 '17 at 9:58

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